Which of the following contains one unpaired electron in the 4p orbitals-

Options

A. `[Cu(NH_3)_2]^+`
B. `[Cu(NH_3)_4]^(2+)`
C. `[Cu(CN)_4]^(3-)`
D. `[Ni(CN_4)]^(2-)`

Answer

To determine which of the given complexes contains one unpaired electron in the 4p orbitals, we will analyze each complex step by step.

### Step 1: Analyze Cu(NH3)2^+
- **Copper's Atomic Number**: 29
- **Electronic Configuration**: The electronic configuration of copper is [Ar] 4s¹ 3d¹⁰.
- **For Cu^+**: When copper loses one electron to form Cu^+, the configuration becomes [Ar] 3d¹⁰ (the 4s electron is removed).
- **Hybridization**: With two NH3 ligands, the coordination number is 2, leading to sp hybridization. This results in a linear geometry.
- **Electron Distribution**: All d electrons are paired (3d¹⁰), and there are no electrons in the 4p orbital.
- **Conclusion**: This complex does not have an unpaired electron in the 4p orbital.

### Step 2: Analyze Cu(NH3)4^2+
- **For Cu^2+**: The charge of +2 means we remove two electrons from the neutral copper configuration. Thus, the configuration becomes [Ar] 3d⁹ (removing one from 4s and one from 3d).
- **Hybridization**: With four NH3 ligands, the coordination number is 4, leading to dsp² hybridization and a square planar geometry.
- **Electron Distribution**: The 3d orbital has 9 electrons, and due to strong field ligands (NH3), one of the d electrons will be promoted to the 4p orbital.
- **Conclusion**: This complex has one unpaired electron in the 4p orbital.

### Step 3: Analyze Cu(CN)4^3-
- **For Cu^1+**: The charge of +1 means that the configuration remains [Ar] 3d¹⁰ (as in the first complex).
- **Hybridization**: With four cyanide ligands, the coordination number is 4, leading to sp³ hybridization.
- **Electron Distribution**: All d electrons are paired (3d¹⁰), and there are no electrons in the 4p orbital.
- **Conclusion**: This complex does not have an unpaired electron in the 4p orbital.

### Step 4: Analyze Ni(CN)4^2-
- **Nickel's Atomic Number**: 28
- **For Ni^2+**: The configuration becomes [Ar] 3d⁸ (removing two electrons from 4s and 3d).
- **Hybridization**: With four cyanide ligands, the coordination number is 4, leading to dsp² hybridization and a square planar geometry.
- **Electron Distribution**: The 3d orbital has 8 electrons. Due to strong field ligands (CN⁻), pairing occurs in the d orbitals, leaving no unpaired electrons.
- **Conclusion**: This complex does not have an unpaired electron in the 4p orbital.

### Final Conclusion:
The only complex that contains one unpaired electron in the 4p orbital is **Cu(NH3)4^2+**.